Equilibrium Process Practice Exam Equilibrium Name (last) (First) Read all questions before you start. Show all work and explain your answers to receive full credit. Report all numerical answers to the proper number of significant figures. By signing your signature above you agree that you have worked alone and neither give nor received help from any source. Keep your eyes on your own paper at all times System Pressure: LENGTH: VOLUME MASS Temperature English: 760 mmhg = 14.7 psi 1 atm = 101.3 KPa 1 ft = 12 in 1 mile = 5280 ft 1 yd = 3 ft 1 gal = 4 qt 1 qt = 57.75 in 3 1 lb = 16 oz 1 ton = 2000lb T F = 1.8 (T C ) + 32 SI- English: 1 atm = 760 torr 1atm = 760 mmhg 1 in = 2.54 cm 1 mi = 1.609 km 1 L = 1.057 qt 1 qt = 0.946 L 1 lb = 453.6 g 1 oz = 28.35 g T C = (T F - 32) / 1.8 Misc. info 1 J = 1 kg m 2 / s 2 1 mole = 6.02 10 23 Density H 2 O: 1.0 g/ml Equilibrium constant K p and K c : K p = K c (RT) Δn (R = 0.08206 L atm / mol K) Quadratic Equation: ax 2 + bx + c = 0 x = b ± b2 4ac 2a
1 What is the general equilibrium constant expression, K eq, for the reversible reaction: 2 A (aq) + 3 B (aq) D C (l) a) K eq = [A] 2 [B] 3 /[C] 0 b) K eq = [A] 2 [B] 3 /[C] 1 c) K eq = [C] 0 / [A] 2 [B] 3 d) K eq = [C] 1 / [A] 2 [B] 3 e) none of the above 2 At 100 C, K p is 2.65 for the equilibrium: SO2Cl2(g) D SO2(g) + Cl2(g). If ΔH rxn = +93.1 kj/mol, what is K p at 200 C? a) K p = 2.65 b) K p < 2.65 c) K p > 2.65 d) none of these e) cannot be determine 3 In which reaction does K p = K c? a) NO (g) + O 2 (g) D N 2 O 3(g) b) N 2 (g) + O 2 (g) D 2 NO (g) c) CaCO 3 (s) D CaO (s) + CO 2 (g) d) N 2 (g) + H 2 O (g) D NO ( g) + H 2(g) e) none 4 Which of the following will shift the reaction to the left (reactant) C 2 H 4 (g) + H 2 O (g D C 2 H 5 OH (g) ΔH = -47.8 kj a) Addition of steam b) removal of ethanol c) increase in pressure d) addition of heat e) none 5 Consider the diagram to the right for: A (g) + B (g) D C (g) ΔH = ( + ) From this reaction coordinate diagram, K eq can be determined to be: a) K eq < 1 b) K eq = 1 c) K eq > 1 d) K eq cannot be determined.
6 Concentration and solubility of gas The solubility of CO2 gas in water is 0.240 g per 100 ml at a pressure of 1.00 atm and 10.0 C. Consider a 100-mL solution that is saturated with CO 2 at 1.00 atm and 10 C. You can assume that the partial pressure of CO 2 is 1.00atm. Answer the following question concerning this solution. i) (Multiple choice) If the pressure is change to 760 mmhg: a) more CO 2 can be dissolve in solution b) The solution is saturated with CO 2 c) The solution will release CO 2 gas d) the temperature of the solution will drop e) none ii) (Multiple choice) If the temperature is raised to 25 C: a) The solution is saturated with CO 2 b) more CO 2 can be dissolve in solution c) The solution will release CO 2 gas d) the pressure of the solution will drop e) none iii) (Multiple choice) If an additional 50-mL of solvent is added : a) The solution is saturated with CO 2 b) the mole fraction, χ, of CO 2 will increase c) The solution will release CO 2 gas d) more CO 2 can be dissolve in solution e) none iv) Calculate the molar concentration (M), molality (m), parts per million concentration (ppm), and mole fraction (χ) of CO 2 at 10.0 C. Assume the density of the solution = 1.00 g/cc M = 0.240g 1.0mol 44.0g 1 100ml 1000ml = 5.45 10 2 1L ppm = 0.240g 100g + 0.240g 106 = 2394ppm m = 0.240g 1.0mol 44.0g )# 1 % $ 100ml 1ml & (* 1000g, +. = 5.45 10 2 * 1g ' 1kg - 0.240g 1mol 44.0g χ = $ 0.240g 1mol ' & ) + 100g 1mol = 9.81 10 4 $ ' & ) % 44.0g ( % 18.0g ( iv) Calculate K p for the above reaction at 10 C. The Mass Action expression is K c = [CO 2(aq) ] /[CO 2(g) ] [ CO 2 (aq) ] K c = [ CO 2 (g) ] = 0.0545 M K = 1.266 p = K c (RT ) 1 1 $ L atm = 1.266 0.08206 mol K 283K ' & ) 1atm % ( mass MWt = P RT # % $ L atm 0.08206 mol K 283K & ( ' = 5.45 10 2 7 Tooth enamel is composed of the mineral hydroxylapatite, Ca 5 (PO 4 ) 3 OH ( = 6.8 10-37 ). By adding fluoride to drinking water, the fluoride can reacts with hydroxylapatite to form a more tooth decay-resistant mineral, called fluorapatitie, Ca 5 (PO 4 ) 3 F (Ksp = 1.0 10-60 ) Calculate the solubility of both hydroxylapatite and fluorapatitie. How many times greater is this mineral more soluble? Ca 5 (PO 4 ) 3 OH! 5Ca +2 + 3PO 4-3 + OH - Ca 5 (PO 4 ) 3 F! 5Ca +2 + 3PO 4-3 + F - = 6.8 10-37 = [Ca +2 ] 5 [PO -3 4 ] 3 [OH - ] 6.8 10-37 = [5s] 5 [3s] 3 [s] 6.8 10-37 = 3125s 5 27s 3 s 6.8 10-37 = 84375s 9 8.06 10-42 = s 9 2.72 10-05 = s = 1.0 10-60 = [Ca +2 ] 5 [PO -3 4 ] 3 [F - ] 1.0 10-60 = [5s] 5 [3s] 3 [s] 1.0 10-60 = 3125s 5 27s 3 s 1.0 10-60 = 84375s 9 1.19 10-65 = s 9 6.11 10-08 = s 2.72 10-5 M / 6.11 10-8 M = 445 Ca 5 (PO 4 ) 3 OH is 445 more soluble than Ca 5 (PO 4 ) 3 F
8 K eq relation to chemical equation As an environmental chemist you are studying catalytic converters and urban smog. (1) N 2 (g) + O 2 (g) D NO (g) K c = 4.8 10-10 (2) NO 2 (g) D NO (g) + O 2 (g) K c = 1.1 10-5 Using the information above, what is K c for the following reaction: (3) 2 NO 2 (g) D N 2 (g) + 2 O 2 (g) K c =?, T = 35 C Rev Rxn1 : 2NO (g) N 2 (g) + O 2 (g) 1 K = K'c Rxn2 : 2NO 2(g) 2NO (g) + O 2 (g) K = K"c Rxn3: 2NO 2(g) 2N 2(g) + 2O 2 (g) Kc = K"c K'c 1.1 10 5 = = 2.29 104 10 4.8 10 1) 2NO (g) D N 2 (g) + O 2 (g) Rev Rxn 1 2) 2NO 2 (g) D 2 NO (g) + O 2 (g) Keep Rxn 2 3) 2 NO 2 (g) D N 2 (g) + 2 O 2 (g) Kc = K c / K c 9 Equilibrium concentration and Keq When 1.00 mol of PCl 5 is introduced into a 5.000 L container at 500K, 75.00 % of the PCl 5 dissociates to give an equilibrium mixture of PCl 5, PCl 3 and Cl 2. PCl 5 (g) D PCl 3 (g) + Cl 2 (g) Calculate [PCl 5, [PCl 3 ] and [Cl 2 ] at equilibrium and K c & K p. PCl 5 (g) PCl 3 (g) + Cl 2 (g) i 0.20 M 0 M 0M Δ - x M + x + x e (.75.20)M x x [ PCl5] eq =.20 - x =.05, x =.15 [ K c = Cl 2 ] [ PCl 3 ] [ =.15 ] [.15 ] =.45 [ PCl 5 ] [.05] K p = K c RT # % $ ( ) 1 =.45.08206 1 L atm mol K 500K & ( = 18.46 ' [ e] [.05 M] [.15] [.15] 10 Determination of Q: Cyclohexane (CH) undergoes a molecular rearrangement in the presence of AlCl3 to form methycyclopentane (MCP) according to the equation CH D MCP: Kc = 0.143 at 25 C for this reaction. Predict the direction the reaction will shift to reach equilibrium if the initial concentrations of the species are [CH] = 0.200 M and [MCP] = 0.100 M.
11 Equilibrium Problem and Q Exactly 0.800 mol of POCl 3 is sealed in a 20-mL flask at 620 K: POCl 3 (g ) D PCl 3 (g) + O 2 (g) The equilibrium constant K c, is 0.090 for this reaction. i) Which statement is true about the above reaction? a) K c = K p b) This reaction is exothermic c) Addition of N 2 result in a shift of the reaction to the left d) none ii) Calculate K p for the above reaction? Kp = Kc(RT) 2 1 L atm =.090(.08206 mol K 620K)1 = 4.57 iii) If the concentration of each specie is [POCl 3 ]=0.00013, [PCl 3 ] =0.0087 [O 2 ]= 0.0078 in which direction will the equilibrium shift? 2 POCl 3 (g ) D 2 PCl 3 (g) + O 2 (g) Balance the Equation!!!! [ ] 2 [ O 2 ] [ [ ] 2 =.0087]2 [.0078] [.00013] 2 = 34.9 35 Q = PCl 3 POCl 3 Q = 35 > Kc Reaction shifts or moves to the left (Reactant) 12 LeChatelier Principle Please complete the following table for the hypothetical reaction: R(g) D 2P(g) ΔH = (- ) Disturbance Affecting. Net direction of Rxn. (Forward, Reverse or none) Effect on K eq. (Increase, decrease or no change) Concentration Increase [reactant] Forward no change Increase [product] Reverse no change Pressure (Vol) Increase P (volume decrease) Reverse no change Decrease P (increase volume) Forward no change Temperature Increase Temp Reverse decrease Decrease Temp Forward increase Catalyst Add catalyst to reactant none no change
13 Fe(NO 3 ) 3 (0.00100 mol) and KSCN (0.200 mol) are added to water to make 1-liter of solution. The red complex ion FeSCN 2+ is produced. Calculate the conc. of Fe 3+ (aq) & FeSCN2+ (aq) at equilibrium, if K f of the FeSCN2+ is 8.9 10 2. Fe +3 + SCN -! FeSCN +2 [ i] 0.00100 0.200 0 [ Δ] - x - x + x [ eq] 0.00100 - x 0.200 - x + x 8.9 10 2 ( 2.00 10 4.201x + x 2 ) = [ x] 0.178 179.89x + 890x 2 = 0 a =890, b = -179.9, c = 0.178 k f = 8.9 10 2 = [ FeSCN +2 ] [ Fe +3 ] SCN - [ ] x = 179.89 ± ( -179.89 )2 4 890 ( ) 2 890 ( )( 0.178) k f = 8.9 10 2 = [ x] 0.00100 - x [ ] [ 0.200 - x] x = 179.89 ± 31726.7321 2 890 ( ) k f = 8.9 10 2 = [ x] 2.00 10 4.201x + x 2 8.9 10 2 ( 2.00 10 4.201x + x 2 ) = [ x] 0.178 179.89x + 890x 2 = 0 x = 179.89 ± 178.12 = 0.20113 or 0.000994386 1780 x = 9.94 10-4 M [ FeSCN +2 ] = 9.94 10-4 M [ Fe +3 ] = 5.614 10 6 M 14 Consider the table below: Analysis for KCl i) What is the temperature in which K 2 Cr 2 O 7 and KCl both have the same solubility? K 2 Cr 2 O 7 T = 59 s = 38 g per 100 ml H2O K 2 Cr 2 O 7 T = 71 s = 49 g per 100 ml H2O ii) What is the solubility (g per 100g H 2 O) at this temperature. See chart below iii) What is the molar solubility for each of these compounds at this temperature See chart below iv) Calculate the for each of these chemicals at this temperature Ksp = [K+]{Cr 2 O 7 -}= (2s) 2 s = 4s 3 Chem MWt s (g/100ml) [s] M T C Ksp K 2 Cr 2 O 7 294.18 49.00 1.67 71 18.48
15 Solution is for Ca 3 (PO 4 ) 2 Look up the value for the salt Ca3(PO4)2 i) What is the molar solubility? ii) What is the solubility express in g per 100 ml solution? iii) What is the molar solubility in 0.010 M K 3 PO 4 solution? Compound Ksp Solubility 1 Ca 3 (PO 4 ) 2 2.0 10-29 2 Mg 3 (PO 4 ) 2 6.3 10-26 iv) iv) Given 1 10-5 M solution of Ca(NO 3 ) 2, how much H 3 PO 4 is necessary to cause a precipitate? Ca 3 (PO 4 ) 2! 3Ca +2 + 2PO 4-3 MW = 310.18 g mol = 2.0 10-29 = [Ca +2 ] 3 [PO 4-3 ] 2 2.0 10-29 = [3s] 3 [2s] 2 2.0 10-29 = 108s 5 1.85 10-31 = s 5 7.14 10-07 M = s solubility g per 100ml H 2 O s = 7.14 10-07 mol L 310.18g.100L mol 100ml s = 2.21 10-05 g per 100cc H 2 O = 2.0 10-29 = [Ca +2 ] 3 [PO 4-3 ] 2 2.0 10-29 = [3s] 3 [0.010M] 2 2.0 10-29 =.0027s 3 7.41 10-27 = s 3 1.95 10-09 = s = 2.0 10-29 = [Ca +2 ] 3 [PO -3 4 ] 2 2.0 10-29 = [1 10-5 ] 3 [2s] 2 2.0 10-29 = 1.0 10-15 4s 2 2.0 10-29 = 4.0 10-15 s 2 5.00 10-15 = s 2 7.07-08 M= s at Equilb, [PO 2-4 ] =2s = 1.414e -7 M, H 3 PO 4 =1.414e -7 M 16 Check mark if you agree or disagree with the statements below and justify for your answer completely. i) Consider a reaction with an K eq ~ 1.0 * 10-55, when the reaction reaches equilibrium, then no product is ever formed. agree or disagree At equilibrium, products and reactants are still exchanging except that the majority of the chemicals in this mixture is the product ii) Initially when a reaction begins, and reactant converts to product but before the reaction reaches equilibrium, the reaction quotient, Q, is decreasing. agree or disagree If the reaction is moving in the forward direction, the Qc < Keq, Qc is therefore increasing to catch-up with Keq iii) For an exothermic reaction, a decrease in temperature leads to an increase in K eq for the reaction at the new temperature. agree or disagree A g P + E, If the temperature decreases, the reaction shifts to the right so that Keq (new) is bigger than Keq (old) 17 Equilibrium Kp & Kc: Nitric oxide and bromine, at initial partial pressures of 98.40 and 41.30 torr, respectively, were allowed to react at 300 K in a seal vessel. At equilibrium, the partial pressure of Br2 = 22.10 torr. 2NO(g) + Br2(g) D 2 NOBr(g) a) Calculate the pressure (in torr) of each species at equilibrium. b) Calculate Kp
18 Equilibrium: Exactly 0.200 mol of PCl5 is sealed in a 2.0-L flask at 620 K.: PCl5 (g) D PCl3 (g) + Cl2 (g) The equilibrium constant Kc, is 0.600 for this reaction. (i) What is a, b and c in the quadratic equation that will give information on the concentration of all species at equilibrium. (ii) What direction will the rxn proceed if the concentrations are: [PCl5]=0.00013, [PCl3] =0.0087 [Cl2 ]= 0.0078? (iii) If the equilibrium constant changes to 1.622 at 800 K, is this an exothermic or endothermic reaction? Explain.
19 LeChatelier The Haber process is used for the production of ammonia according to the reaction: N2(g) + H2(g) NH3(g) at 127 C, the K c =3.8 10 4 L 2 mol -2 at 500 C, the K c =6.0 10-2 L 2 mol -2 i) The Haber reaction is: a) An exothermic process b) An endothermic process c) is an adiabatic process d) none ii) The production of ammonia can be increase by: a) increasing the temperature but decreasing the pressure b) decreasing the temperature and pressure c) increasing the pressure but decreasing the temperature d) increasing the pressure and temperature Consider the graph for the Haber reaction. The reaction takes place in a 1-Liter vessel. The initial concentration of [H 2 ] is 12.0 molar and that of [N 2 ] is 4.0 molar. iii) Sketch the change of the concentration of N 2 from 0 to 3.5 min. See Diagram iv) At 2 minutes, Q is: a) increasing b) decreasing c) not changing d ) cannot be determine v) At 5 minutes, Q is: a) increasing b) decreasing c) not changing d ) cannot be determine vi) Calculate the equilibrium constant at the 4.0 minute mark. Q = [ NH 3 ] 2 [ [ H 2 ] 3 [ N 2 ] = 8.0] 2 [ 10.0] 3 [ 2.4] = 0.027.03 v) Calculate the equilibrium constant at the 8.0 minute mark. Q = [ NH 3 ] 2 [ [ H 2 ] 3 [ N 2 ] = 9.0] 2 [ 11.0] 3 [ 2.0] = 0.0304.03 20 Heterogeneous Equilibrium: 10.0 ml of 7.40 10-4 M HCl is added to 10.0 ml silver nitrate solution according to the equation below. For a precipitation to occur, what is the initial concentration of silver nitrate ( or Ag+ )? AgNO3(aq) + HCl(aq) D AgCl(s) + HNO3(aq) Keq = 5.56 109 First write the net ionic equation for the reaction above, then remember (Ksp = 1/Keq ) [Cl - ] = 7.40 10 4 10.0ml 20.0ml = 3.7 10 4 ; = 1 K eq = 1.80 10 10 AgCl (s) Ag + (aq) + Cl - (aq) = 1.80 10 10 [i] Excess 0 3.7 10 4 Δ - s + s + s [e] Excess + s + 3.7 10 4 + s 1.80 10 10 = s (3.7 10 4 ); s = 4.86 * 10-7 [Ag + ] = 4.86 * 10-7 = [AgNO 3 ]